Question

if the zeroes of the polynomial x3 -3x2+x+1 and a-b,a,a+b .find 'a' and 'b'

Answer 1

Answer:  The required value of a is 1 and that of b is ±√2.

Step-by-step explanation:  Given that the zeroes of the following polynomial are a-b, a and a+b :

$p(x)=x^3-3x^2+x+1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to find the values of a and b.

We know that if p, q and r are the zeroes of a polynomial $f(y)=y^3+sy^2+ts+h,$ then

$p+q+r=-s,\\\\pq+qr+rp=t,\\\\pqr=-h.$

So, for polynomial (i), we get

$(a-b)+a+(a+b)=-(-3)\\\\\Rightarrow 3a=3\\\\\Rightarrow a=1$

and

$(a-b)a+a(a+b)+(a+b)(a-b)=1\\\\\Rightarrow (1-b)1+1(1+b)+(1+b)(1-b)=1\\\\\Rightarrow 1-b+1+b+1-b^2=1\\\\\Rightarrow b^2=2\\\\\Rightarrow b=\pm\sqrt{2}~~~~~~~~~~~~~~~~~~[\textup{taking square root on both sides}]$

Thus, the required value of a is 1 and that of b is ±√2.

Answer 2

Answer:

a = 1

b = ±√2

Step-by-step explanation:

$p(x) = x^3 - 3x^2 + x + 1\\\\\text{A standard cubic equation is of the form ax^3 + bx^2 + cx + d }\\\\\text{On comparing the two equations, we get}\\\\a = 1\\\\b = -3\\\\c= 1\\\\\\d=1\\\\\text{Given zeroes are:}\\\\\alpha = a-b\\\\\beta = a\\\\\gamma = a + b\\\\\text{We know that in a cubic polynomial,}\\\\\text{sum of zeroes = -\dfrac{b}{a} }\\\\\\\implies \alpha + \beta + \gamma = -\dfrac{(-3)}{1}\\\\\\\implies a - b + a + a + b = 3\\\\\implies 3a = 3$

$\implies a = 3 \div 3 = 1\\\\\implies \boxed{\underline{\boxed{\bold{a = 1}}}}\\\\\\\text{We also know that in a cubic polynomial,}\\\\\text{Product of zeroes = -\dfrac{c}{a} }\\\\\\\implies \alpha \times \beta \times \gamma = -\dfrac{c}{a}\\\\\\\implies (a-b)(a)(a+b) = -\dfrac{1}{1}\\\\\\\text{Put a = 1}\\\\(1-b)(1)(1+b) = -1\\\\\implies 1^2 - b^2 = -1\\\\\implies 1 - b^2 = -1\\\\\implies b^2 = 1 + 1\\\\\implies b^2 = 2$

$\implies \boxed{\underline{\boxed{\bold{b = \pm \sqrt{2}}}}}$